WAEC SSCE - General Mathematics - 1994 (Essay)

Bayanin Amsa

(a) Solve \(\log_{10}(3x-1) - \log_{10}2 = 3\)

Combine the logs using \(\log a - \log b = \log\frac{a}{b}\):

\[\log_{10}\left(\frac{3x-1}{2}\right) = 3\]

Rewrite in index form (\(\log_{10}N = 3 \Rightarrow N = 10^{3}\)):

\[\frac{3x-1}{2} = 1000\]

\[3x-1 = 2000 \;\Rightarrow\; 3x = 2001 \;\Rightarrow\; x = 667\]

(b) Evaluate \(\sqrt{\dfrac{0.897\times 3.536}{0.00249}}\) to 3 s.f. using logarithm tables

Number	Logarithm
0.897	\(\bar{1}.9528\)
3.536	\(0.5485\)
Numerator sum	\(0.5013\)
0.00249	\(\bar{3}.3963\)

Divide (subtract the log of the denominator):

\[0.5013 - \bar{3}.3963 = 0.5013 - (-2.6037) = 3.1050\]

Take the square root (divide the log by 2):

\[\tfrac{1}{2}\times 3.1050 = 1.5525\]

Antilog of \(1.5525\) gives \(3.570\times 10^{1}\).

\[\sqrt{\frac{0.897\times 3.536}{0.00249}} \approx 35.7\]

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Bayanin Amsa

(a) Divide \(11111111_{two}\) by \(101_{two}\)

We use binary long division. Divisor is \(101_{two}\).

         1 1 0 0 1 1
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1 0 1 ) 1 1 1 1 1 1 1 1
        1 0 1
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            0 0 1 1
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                    0 0

The quotient is \(11111111_{two} \div 101_{two} = 110011_{two}\), remainder \(0\).

Check in base ten: \(11111111_{two}=255\), \(101_{two}=5\), and \(255 \div 5 = 51 = 110011_{two}\). Correct.

(b) Sector: radius 6 cm, angle 105°, \(\pi=\tfrac{22}{7}\)

(i) Perimeter. First find the arc length.

\[\text{Arc} = \frac{\theta}{360}\times 2\pi r = \frac{105}{360}\times 2\times\frac{22}{7}\times 6 = \frac{7}{24}\times\frac{264}{7} = 11\text{ cm}\]

Perimeter = arc + the two straight radii:

\[P = 11 + 2(6) = 23\text{ cm}\]

(ii) Area.

\[A = \frac{\theta}{360}\times \pi r^{2} = \frac{105}{360}\times\frac{22}{7}\times 6^{2} = \frac{7}{24}\times\frac{792}{7} = 33\text{ cm}^{2}\]

Bayanin Amsa

(a) Cumulative frequency table. Ages are given to the nearest 5 years, so the class boundaries are \(7.5\text{-}12.5,\ 12.5\text{-}17.5,\) etc.

Age (nearest 5 yr)	Upper boundary	No. of people	Cumulative freq.
10	12.5	8	8
15	17.5	19	27
20	22.5	10	37
25	27.5	7	44
30	32.5	6	50

(b) Plot cumulative frequency against upper class boundary and join with a smooth curve (ogive).

(c)(i) Median age. At \(\frac{N}{2} = 25\), reading across from the ogive:

\[\text{Median} = 12.5 + \frac{25 - 8}{19}\times 5 \approx 17 \text{ years}\]

(c)(ii) At most 15 years. Reading up from \(x = 15\): about 18 people.

(c)(iii) Between 20 and 25 years. Reading the ogive at 25 and at 20 and subtracting:

\[\approx 40.5 - 32 \approx 8 \text{ people}\]

Bayanin Amsa

Bayanin Amsa

(a) Speed along the parallel of latitude

Both points are on latitude \(32^\circ\)N. The difference in longitude is

\[47^\circ + 25^\circ = 72^\circ\]

Distance along a parallel of latitude \(\phi\):

\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]

\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]

Using \(\cos32^\circ = 0.8480\):

\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]

Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.

Speed \(\approx 620\) km/h.

(b) Height of the kite

Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).

From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).

From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).

Substitute \(d = h\):

\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]

\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]

Height \(\approx 35\) m.

Bayanin Amsa

(a) Use the difference of two squares, \(a^2 - b^2 = (a-b)(a+b)\):

\(17.57^2 - 12.43^2 = (17.57 - 12.43)(17.57 + 12.43)\)

\(= (5.14)(30.00) = \mathbf{154.2}\)

(b) Theorem: Angles in the same segment of a circle are equal.

Given: A circle with centre \(O\); a chord \(AB\); points \(P\) and \(Q\) on the major arc (the same segment), so \(\angle APB\) and \(\angle AQB\) both stand on chord \(AB\).

To prove: \(\angle APB = \angle AQB\).

Construction: Join \(OA\) and \(OB\).

Proof: By the theorem that the angle subtended by an arc at the centre is twice the angle it subtends at the circumference (on the same arc):

\(\angle AOB = 2\,\angle APB \quad\text{and}\quad \angle AOB = 2\,\angle AQB\)

Therefore \(2\,\angle APB = 2\,\angle AQB\), giving \(\angle APB = \angle AQB\). Hence angles in the same segment are equal. \(\blacksquare\)

Bayanin Amsa

Let the horizontal distance between the building and the foot \(F\) of the tower be \(x\), and the tower height be \(H\). \(P\) is at the foot of the building, \(Q\) is its top, \(20\text{ m}\) directly above \(P\).

From \(P\): \(\tan 65° = \dfrac{H}{x}\), so \(H = x\tan 65°\).

From \(Q\) (height 20 m, same horizontal distance \(x\)): \(\tan 25° = \dfrac{H - 20}{x}\), so \(H - 20 = x\tan 25°\).

Subtracting: \(20 = x(\tan 65° - \tan 25°) = x(2.1445 - 0.4663) = 1.6782x\).

\(x = \dfrac{20}{1.6782} = 11.92\text{ m}\).

(a) Distance PT: \(PT\) is the line of sight from \(P\) to \(T\), at elevation \(65°\):

\(PT = \dfrac{x}{\cos 65°} = \dfrac{11.92}{0.4226} = 28.20 \approx \mathbf{28.2\text{ m}}\) (3 s.f.).

(b) Height of tower: \(H = x\tan 65° = 11.92 \times 2.1445 = 25.56 \approx \mathbf{25.6\text{ m}}\) (3 s.f.).

Bayanin Amsa

(a) Arithmetic Progression

Let the first term be \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).

Fourth term: \(a + 3d = 37\).

The 6th term is 12 more than the 4th term: \((a+5d) = 37 + 12 = 49\), so \(a + 5d = 49\).

Subtract the equations:

\[(a+5d)-(a+3d) = 49-37 \;\Rightarrow\; 2d = 12 \;\Rightarrow\; d = 6\]

Then \(a = 37 - 3(6) = 19\).

First term \(= 19\). Seventh term:

\[a + 6d = 19 + 6(6) = 55\]

Seventh term \(= 55\).

(b) Sets with \(P=\{1,2,3,4\}\) and \(Q=\{3,5,6\}\).

(i) \(P\cap Q = \{3\}\)

(ii) \(P\cup Q = \{1,2,3,4,5,6\}\)

(iii) \((P\cap Q)\cup Q = \{3\}\cup\{3,5,6\} = \{3,5,6\}\)

(iv) \((P\cap Q)\cup P = \{3\}\cup\{1,2,3,4\} = \{1,2,3,4\}\)

Bayanin Amsa

Setup. Bag A: 12 white, 8 black (20 balls). Bag B: 10 white, 15 black (25 balls). Two balls are drawn from each bag without replacement.

Useful probabilities per bag

Bag A, two black: \(\dfrac{8}{20}\times\dfrac{7}{19} = \dfrac{14}{95}\)

Bag A, exactly one white: \(2\times\dfrac{12}{20}\times\dfrac{8}{19} = \dfrac{48}{95}\)

Bag B, two black: \(\dfrac{15}{25}\times\dfrac{14}{24} = \dfrac{7}{20}\)

Bag B, exactly one white: \(2\times\dfrac{10}{25}\times\dfrac{15}{24} = \dfrac{1}{2}\)

(a) All four balls black

Both bags must give two blacks:

\[P = \frac{14}{95}\times\frac{7}{20} = \frac{98}{1900} = \frac{49}{950} \approx 0.052\]

(b) Exactly one of the four balls is white

The single white must come from one bag while the other bag yields two blacks.

White from A, two black from B: \(\dfrac{48}{95}\times\dfrac{7}{20} = \dfrac{84}{475}\)

Two black from A, white from B: \(\dfrac{14}{95}\times\dfrac{1}{2} = \dfrac{7}{95} = \dfrac{35}{475}\)

Add the mutually exclusive cases:

\[P = \frac{84}{475} + \frac{35}{475} = \frac{119}{475} \approx 0.251\]

Bayanin Amsa

This is a discrete frequency distribution, so each score \(x\) already has its frequency \(f\). Build an \(fx\) column, add the columns, then apply the mean formula.

Scores \(x\)	15	25	35	45	55	65	75	Total
Freq \(f\)	1	4	12	24	18	8	3	70
\(fx\)	15	100	420	1080	990	520	225	3350

(a) Mean mark.

\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3350}{70} = 47.857\ldots\]

Correct to 3 significant figures, the mean mark is \(\mathbf{47.9}\).

(b)(i) Mode. The mode is the score with the highest frequency. The largest frequency is \(24\), which belongs to the score \(45\), so the mode is \(\mathbf{45}\).

(b)(ii) Range.

\[\text{Range} = \text{highest score} - \text{lowest score} = 75 - 15 = \mathbf{60}.\]

Examination note: multiply every score by its own frequency before adding, and divide by \(\sum f = 70\), not by the number of different scores (7). Dividing by 7 is the usual slip that gives a wrong mean near \(46\text{-}48\) instead of the correct \(47.9\).

Bayanin Amsa

(a) Express everything as powers of 3: \(9 = 3^2\), \(27 = 3^3\).

\(3\times 9^{1+x} = 3^1 \times 3^{2(1+x)} = 3^{1 + 2 + 2x} = 3^{3 + 2x}\)

\(27^{-x} = 3^{-3x}\)

Equating indices: \(3 + 2x = -3x \ \Rightarrow\ 5x = -3 \ \Rightarrow\ \mathbf{x = -\dfrac{3}{5}}\).

(b) Use \(\log\sqrt{a} = \tfrac{1}{2}\log a\) and combine:

\(\log_{10}\sqrt{35} + \log_{10}\sqrt{2} - \log_{10}\sqrt{7} = \log_{10}\!\left(\dfrac{\sqrt{35}\times\sqrt{2}}{\sqrt{7}}\right)\)

\(= \log_{10}\sqrt{\dfrac{35\times 2}{7}} = \log_{10}\sqrt{10} = \log_{10} 10^{1/2} = \dfrac{1}{2}\)

Value \(= \mathbf{\dfrac{1}{2}}\).

Bayanin Amsa

(a) Completing the table for \(y = 2x^{2} - 9x - 1\).

• \(x=-1:\ 2(1)-9(-1)-1 = 2+9-1 = 10\)
• \(x=3:\ 2(9)-9(3)-1 = 18-27-1 = -10\)
• \(x=4:\ 2(16)-9(4)-1 = 32-36-1 = -5\)
• \(x=5:\ 2(25)-9(5)-1 = 50-45-1 = 4\)

\(x\)	-1	0	1	2	3	4	5	6
\(y\)	10	-1	-8	-11	-10	-5	4	17

(b) Plot the points and join with a smooth parabola.

(c)(i) Roots of \(2x^{2} - 9x = 4\). Since \(y = 2x^{2}-9x-1\), rewrite \(2x^{2}-9x-4=0\) as \(2x^{2}-9x-1 = 3\). Draw the line \(y = 3\) and read where it cuts the curve:

\[x = \frac{9 \pm \sqrt{81+32}}{4} = \frac{9 \pm 10.63}{4} \;\Rightarrow\; x \approx -0.4 \text{ or } 4.9\]

(c)(ii) Gradient at \(x = 3\). Draw a tangent at \(x=3\); its slope is

\[\frac{dy}{dx} = 4x - 9 = 4(3) - 9 = \mathbf{3}\]